Question

Three capacitors each having capacitance $C=2\mu \text{F}$ are connected with a battery of e.m.f. $30\text{V}$ as shown in the figure. When the switch $S$ is closed, the extra amount of charge flown through the battery is

• A. $20\mu \text{C}$
• B. $24\mu \text{C}$
• C. $10\mu \text{C}$
• D. $40\mu \text{C}$

Answer 1

The correct option is A $20\mu \text{C}$

Before $S$ is closed,
$\Rightarrow Q_i=\frac{2\text{C}}{3}\times 30$
$\Rightarrow Q_i=40\mu \text{C}$
After $S$ is closed, 1and 2 are shorted.


$\Rightarrow Q_f=C\times 30$
$=60\mu \text{C}$
The charge that flows through the battery is,
$\Delta Q=Q_f-Q_i$
$=60-40$
$=20\mu \text{C}$